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IB Mathematics: Applications and Interpretation HL · AHL 2: Further Derivatives · AHL 2.10

Mixed practice: stationary points, concavity, parameters and models Problem set A One question at a time

Try each question on paper first. Open the working a line at a time, then say whether you had it.

Skills

  1. Sketch a graph of \(y=f(x)\) where \(f'(x)<0\) for all x, \(f''(x)<0\) for \(x<1\) and \(f''(x)>0\) for \(x>1\). Mark the point where the bend changes.

    • \(f'<0\): the curve goes down from left to right everywhere.
    • \(f''<0\) for \(x<1\): the gradient is falling, so the curve is steeply downward and getting steeper. It bends down.
    • \(f''>0\) for \(x>1\): the gradient is rising towards 0 (still negative), so the curve flattens out. It bends up.
    • \(f''\) changes sign at \(x=1\): mark a point of inflection there. The gradient is at its most negative at that point.

  2. The curve \(y=x^3+px^2-2x+9\) is concave up for \(x>-1\) only.

    1. (a) Find the value of p.

      • \(y''=2 \left(p + 3 x\right)\)
      • \(y''=0\) at \(x=-1\): \(2 p - 6=0\)
      • \(p=3\)

    2. (b) What can you say about the curve at \(x=-1\)?

      • \(y''=6 x + 6\), which is 0 at \(x=-1\) and changes sign there.
      • The curve has a point of inflection at \(x=-1\).

  3. The function \(f(x)=x^n\) has \(f''(1)=20\). Find the possible values of n.

    • \(f'=nx^{n-1}\)
    • \(f''=n(n-1)x^{n-2}\)
    • \(f''(1)=n(n-1)=20\), so \(n^2-n-20=0\)
    • \((n-5)(n+4)=0\)
    • \(n=5\) or \(n=-4\)

  4. A curve has a minimum point. Find the values of b and c in each case.

    1. (a) \(y=x^2+bx+c\) has a minimum at \((-2,5)\).

      • \(y'=2x+b=0\) at \(x=-2\): \(-4+b=0\), so \(b=4\)
      • \(y=5\) at \(x=-2\): \(4+(4)(-2)+c=5\)
      • \(c=9\)

    2. (b) \(y=x^4+bx+c\) has a minimum at \((2,-10)\).

      • \(y'=4x^3+b=0\) at \(x=2\): \(32+b=0\), so \(b=-32\)
      • \(y=-10\) at \(x=2\): \(16+(-32)(2)+c=-10\)
      • \(c=38\)

  5. Find the stationary points of each curve and classify them.

    1. (a) \(y=2x-5\sqrt x\), for \(x>0\). Give exact values.

      • \(y'=2 - \frac{5}{2 \sqrt{x}}=0\), so \(\sqrt x=\frac{5}{4}\) and \(x=\frac{25}{16}\)
      • \(y''=\frac{5}{4 x^{\frac{3}{2}}}>0\) for all \(x>0\), so it is a minimum
      • \(y=- \frac{25}{8}\)

    2. (b) \(y=\dfrac{x}{x^2+4}\). Give exact values.

      • \(y'=\dfrac{4-x^2}{(x^2+4)^2}=0\) when \(x^2=4\), so \(x=\pm2\)
      • \(y''=\frac{2 x \left(x^{2} - 12\right)}{\left(x^{2} + 4\right)^{3}}\): \(y''(-2)=\frac{1}{16}>0\) and \(y''(2)=- \frac{1}{16}<0\)
      • Minimum at \(x=-2\), maximum at \(x=2\)

    3. (c) \(y=x+2\cos x\), for \(0<x<2\pi\). Give exact values.

      • \(y'=1 - 2 \sin{\left(x \right)}=0\), so \(x\) is \(\frac{\pi}{6}\) or \(\frac{5 \pi}{6}\)
      • \(y''=- 2 \cos{\left(x \right)}\): \(y''(\frac{\pi}{6})=- \sqrt{3}<0\) and \(y''(\frac{5 \pi}{6})=\sqrt{3}>0\)
      • Maximum at \(x=\frac{\pi}{6}\), minimum at \(x=\frac{5 \pi}{6}\)

  6. Show each of these results.

    1. (a) The graph of \(y=3x^2+bx+c\) is concave up for every value of x.

      • \(y'=6x+b\)
      • \(y''=6\)
      • \(6>0\) for all x, so the graph is always concave up.

    2. (b) If \(y=3\sin 2x+5\cos 2x\), then \(\dfrac{d^2y}{dx^2}=-4y\).

      • \(y'=- 10 \sin{\left(2 x \right)} + 6 \cos{\left(2 x \right)}\)
      • \(y''=- 4 \left(3 \sin{\left(2 x \right)} + 5 \cos{\left(2 x \right)}\right)\)
      • \(=-4\left(3\sin 2x+5\cos 2x\right)=-4y\)

    3. (c) The graph of \(y=\ln(a+2x)\) is always concave down.

      • \(y'=\dfrac{2}{a+2x}=2(a+2x)^{-1}\)
      • \(y''=-2\cdot2(a+2x)^{-2}=\dfrac{-4}{(a+2x)^2}\)
      • The denominator is a square, so it is positive. Then \(y''<0\) wherever the graph exists: always concave down.

  7. The curve is \(y=x^3-9x^2+15x+4\).

    1. (a) Find the interval in which the gradient is decreasing.

      • \(y'=3 x^{2} - 18 x + 15\)
      • \(y''=6 \left(x - 3\right)\)
      • The gradient is decreasing when \(y''<0\): \(6 \left(x - 3\right)<0\), so \(x<3\)

    2. (b) Hence find the coordinates of the point of inflection.

      • \(y''=0\) at \(x=3\), and \(y''\) changes from negative to positive there
      • \(y(3)=-5\)
      • Point of inflection \((3,-5)\)

  8. The curve is \(y=x^4\ln x\), for \(x>0\).

    1. (a) Find the x-coordinate of the point where \(y''=2\). Use the calculator to solve the equation.

      • \(y'=4x^3\ln x+x^{3}\)
      • \(y''=x^{2} \left(12 L + 7\right)\) with \(L=\ln x\), so \(y''=x^2(12\ln x+7)\)
      • From the calculator: \(x^{2}(12\ln x+7)=2\) gives \(x=0.750\)

    2. (b) The curve has a point of inflection. Find its x-coordinate.

      • \(y''=0\): \(12\ln x+7=0\) (as \(x>0\))
      • \(\ln x=-\dfrac{7}{12}\), so \(x=e^{-7/12}=0.558\)
      • y″ changes sign here, so it is a point of inflection.

  9. The function f is defined by \(f(x)=\dfrac{\ln x}{x^3}\), for \(x>0\).

    1. (a) Find \(f'(x)\).

      • Quotient rule: \(f'=\dfrac{x^3\cdot\frac1x-3x^2\ln x}{x^{6}}=\dfrac{x^{2}(1-3\ln x)}{x^{6}}\)
      • \(f'(x)=\dfrac{1-3\ln x}{x^4}\)

    2. (b) Find the exact coordinates of the stationary point.

      • \(f'=0\): \(1-3\ln x=0\), so \(\ln x=\dfrac13\) and \(x=e^{1/3}\)
      • \(x^3=e\), so \(f=\dfrac{1/3}{e}=\dfrac1{3e}\)

    3. (c) Show that \(f''(x)=\dfrac{12\ln x-7}{x^5}\).

      • Quotient rule: \(f''=\dfrac{-\frac3x\cdot x^4-4x^3(1-3\ln x)}{x^{8}}\)
      • \(=\dfrac{-3x^3-4x^3+12x^3\ln x}{x^{8}}\)
      • \(=\dfrac{12\ln x-7}{x^5}\)

    4. (d) Hence classify the stationary point.

      • At \(x=e^{1/3}\), \(\ln x=\dfrac13\), so the numerator is \(-3<0\)
      • The denominator is positive, so \(f''<0\): a local maximum.

In context

  1. A bakery makes x hundred loaves a day, where \(x>0\). The daily profit, in hundreds of dollars, is modelled by \(P=-x^3+12x^2-21x-30\).

    1. (a) Find the value of x that gives the greatest profit. Show that it is a maximum.

      • \(P'=- 3 x^{2} + 24 x - 21=0\): \(- 3 \left(x - 7\right) \left(x - 1\right)=0\)
      • \(x=1\) or \(x=7\)
      • \(P''=6 \left(4 - x\right)\): \(P''(7)=-18<0\), so \(x=7\) is a maximum

    2. (b) Find the greatest profit.

      • \(P(7)=68\) hundred dollars

  2. Water runs into a garden tank. The volume of water, V litres, after t hours is modelled by \(V=t^3-6t^2+17t\), for \(t\ge0\). Find the time at which the tank is filling most slowly, and state the minimum rate of filling.

    • The rate of filling is \(V'=3 t^{2} - 12 t + 17\)
    • Minimise \(V'\): \(V''=6 \left(t - 2\right)=0\) gives \(t=2\)
    • \(V'''=6>0\), so \(V'\) has a minimum at \(t=2\)
    • \(V'(2)=5\) litres per hour

  3. The number of seabirds on an island, P thousand, is modelled by \(P=\dfrac{12t^2}{9+t^2}\), where t is the time in years since monitoring began.

    1. (a) Find \(\dfrac{dP}{dt}\) and use it to find the time at which the number is increasing most quickly. Give your answer to 3 significant figures.

      • Quotient rule: \(\dfrac{dP}{dt}=\dfrac{216t}{(9+t^2)^2}\)
      • Greatest rate: \(P''=- \frac{648 \left(t^{2} - 3\right)}{\left(t^{2} + 9\right)^{3}}=0\)
      • \(9-3t^2=0\), so \(t=\sqrt{3}=1.73\)
      • \(P''\) changes from positive to negative here, so \(P'\) is a maximum.

    2. (b) Find the number at this time and the greatest rate of increase.

      • \(P=12t^2/(9+t^2)\) at \(t^2=3\): \(P=3.00\) thousand
      • Rate: \(P'=2.60\) thousand per year

Stretch

  1. Find the stationary points of the curve \(y=px+\dfrac{4}{x}\), where \(p>0\), and classify them. Give your answers in terms of p.

    • \(y'=p-\dfrac{4}{x^2}=0\), so \(x^2=\dfrac{4}{p}\) and \(x=\pm\dfrac{2}{\sqrt p}\)
    • \(y=\pm\left(\sqrt p\cdot2+2\sqrt p\right)=\pm4\sqrt p\)
    • \(y''=\dfrac{8}{x^3}\): positive for \(x>0\), negative for \(x<0\)
    • \(x=\dfrac{2}{\sqrt p}\) is a minimum; \(x=-\dfrac{2}{\sqrt p}\) is a maximum

  2. A cuboid has a square base of side x cm and height \((a-2x)\) cm, where a is a positive constant and \(0<x<\dfrac{a}{2}\).

    1. (a) Find the value of x, in terms of a, that gives the greatest volume. Justify that it is a maximum.

      • \(V=x^2(a-2x)=a x^{2} - 2 x^{3}\)
      • \(V'=2 a x - 6 x^{2}=- 2 x \left(- a + 3 x\right)=0\), so \(x=\frac{a}{3}\) (\(x=0\) is not allowed)
      • \(V''=2 \left(a - 6 x\right)\), and \(V''\left(\frac{a}{3}\right)=- 2 a<0\) because \(a>0\)
      • So the volume is a maximum.

    2. (b) Find the greatest volume in terms of a.

      • \(V\left(\frac{a}{3}\right)=\frac{a^{3}}{27}\)

  3. Answer both parts.

    1. (a) Find the range of \(f(x)=e^{2x}-ax\), where \(a>0\), in terms of a.

      • \(f'=2e^{2x}-a=0\), so \(e^{2x}=\dfrac{a}{2}\) and \(x=\dfrac12\ln\dfrac{a}{2}\)
      • \(f''=4e^{2x}>0\), so this is the minimum
      • \(f_{\min}=\dfrac{a}{2}-\dfrac{a}{2}\ln\dfrac{a}{2}\)
      • Range: \(f(x)\ge\dfrac{a}{2}\left(1-\ln\dfrac{a}{2}\right)\)

    2. (b) A model for the potential energy U joules of a particle at position x metres is \(U=e^{2x}+e^{-3x}\). Find the exact minimum value of U.

      • \(U'=2e^{2x}-3e^{-3x}=0\), so \(e^{5x}=\dfrac{3}{2}\) and \(x=\dfrac{1}{5}\ln\dfrac{3}{2}\)
      • \(e^{2x}=\left(\dfrac{3}{2}\right)^{2/5}\) and \(e^{-3x}=\left(\dfrac{3}{2}\right)^{-3/5}=\frac{2}{3}\left(\dfrac{3}{2}\right)^{2/5}\)
      • \(U''=4e^{2x}+9e^{-3x}>0\): a minimum
      • \(U_{\min}=\left(1+\frac{2}{3}\right)\left(\dfrac{3}{2}\right)^{2/5}=\dfrac{5}{3}\left(\dfrac{3}{2}\right)^{2/5}=1.96\) J

  4. Answer both parts.

    1. (a) The curve \(y=x^3-3x^2+bx+2\) has a positive gradient at every point (it is strictly increasing and never flat). Find the condition on b.

      • \(y'=b + 3 x^{2} - 6 x\)
      • A positive gradient at every point means \(y'>0\) for all x, so \(y'=0\) must have no real roots: \(\Delta<0\)
      • \(\Delta=(-6)^2-4(3)(b)=36 - 12 b<0\)
      • \(b>3\)

    2. (b) The curve \(y=ax^3-bx^2\), where a and b are positive, is increasing for \(x<0\) and for \(x>6\). Evaluate \(\dfrac{b}{a}\).

      • \(y'=3ax^2-2bx=x(3ax-2b)\)
      • \(y'\) changes sign at \(x=0\) and at \(x=\dfrac{2b}{3a}\), which is the \(x=6\) boundary
      • \(\dfrac{2b}{3a}=6\), so \(\dfrac{b}{a}=9\)

  5. A curve has \(y''=6x+4\). Its gradient at \(x=0\) is −2 and it passes through the point \((1,5)\). Find y in terms of x.

    • \(y'=\int(6x+4)\,dx=3 x^{2} + 4 x+C\)
    • \(y'(0)=-2\) gives \(C=-2\), so \(y'=3 x^{2} + 4 x - 2\)
    • \(y=x^{3} + 2 x^{2} - 2 x+D\)
    • \(y(1)=5\) gives \(D=4\)
    • \(y=x^{3} + 2 x^{2} - 2 x + 4\)

  6. The graph shows \(y=f'(x)\). It crosses the x-axis at \(x=-2\) and touches the x-axis at \(x=1\).

    −3−2−112−15−10−5510xyy = f′(x)

    1. (a) Write down the x-coordinate of each local minimum of \(f\), and of each local maximum of \(f\).

      • Local extrema of f are where \(f'\) crosses the x-axis.
      • At \(x=-2\), \(f'\) goes from negative to positive: local minimum.
      • At \(x=1\), \(f'\) touches the axis and stays positive: not a maximum or minimum.

    2. (b) Write down the x-coordinate of each point of inflection of \(f\).

      • Inflections of f are where \(f'\) has a local maximum or minimum (f″ changes sign).
      • \(f'\) has a local maximum at \(x=-1\) and a local minimum at \(x=1\).

    3. (c) Which of these points of inflection are stationary points of inflection? Explain.

      • At \(x=1\), \(f'=0\) as well as \(f''=0\): a stationary point of inflection.
      • At \(x=-1\), \(f'=4\), not 0: not stationary.

jmaths.xyz · AI HL · AHL 2.10 Mixed practice: stationary points, concavity and models